Problem: Write the equation for a parabola with a focus at $(2,2)$ and a directrix at $x=8$. $x=$
The strategy A parabola is defined as the set of all points that are the same distance away from a point (the focus) and a line (the directrix). Let $(x,y)$ be a point on the parabola. Then the distance between $(x,y)$ and the focus, $(2,2)$, is equal to the distance between $(x,y)$ and the directrix, $x=8$. Once we find these distances, we can equate them in order to derive the equation of our parabola. Finding the distances from $(x,y)$ to the focus and the directrix The distance between $(x,y)$ and $(2,2)$ is $\sqrt{(x-2)^2+(y-2)^2}$. [How did we find that?] Similarly, the distance between $(x,y)$ and the line $x=8$ is $\sqrt{(x-8)^2}$. [How did we know that?] Deriving the formula by equating the distances $\begin{aligned} \sqrt{(x-8)^2} &= \sqrt{(x-2)^2+(y-2)^2} \\\\ (x-8)^2 &= (x-2)^2+(y-2)^2 \\\\ {x^2}-16x{+64} &= {x^2}{-4x}+4+(y-2)^2\\\\ -16x{+4x}&=(y-2)^2+4{-64} \\\\ -12x&=(y-2)^2-60 \\\\ x&=-\dfrac{(y-2)^2}{12}+5\end{aligned}$ The answer The equation of our parabola is $x=-\dfrac{(y-2)^2}{12}+5$. Here is the graph of our parabola. As expected, the distance between a point on the parabola, $(x,y)$, and the focus is the same as the distance between $(x,y)$ and the directrix. ${1}$ ${2}$ ${3}$ ${4}$ ${5}$ ${6}$ ${7}$ ${8}$ ${9}$ ${\llap{-}2}$ ${\llap{-}3}$ ${\llap{-}4}$ ${\llap{-}5}$ ${\llap{-}6}$ ${\llap{-}7}$ ${\llap{-}8}$ ${\llap{-}9}$ ${1}$ ${2}$ ${3}$ ${4}$ ${5}$ ${6}$ ${7}$ ${8}$ ${9}$ ${10}$ ${11}$ ${\llap{-}2}$ ${\llap{-}3}$ ${\llap{-}4}$ ${\llap{-}5}$ ${\llap{-}6}$ ${\llap{-}7}$ $y$ $x$ ${(x,y)}$